Law of Total Probability
Suppose event $A$ can occur given the occurrence of one of the events $B_1, B_2, …, B_n$, which form a complete group of events. Suppose the probabilities of these events $P(B_1), P(B_2), …, P(B_n)$ are known, as well as the conditional probabilities $P(A\mid B_1), P(A\mid B_2), …, P(A\mid B_n)$ of event $A$ given each of $B_1, …, B_n$.
Theorem. The probability of event $A$, which can occur only given the occurrence of one of the mutually exclusive events $B_1, B_2, …, B_n$ forming a complete group, is
$P(A) = P(B_1) P(A\mid B_1) + P(B_2) P(A\mid B_2) + … + P(B_n) P(A\mid B_n)$
This formula is called the law of total probability (it follows from the definition of conditional probability).
Proof
Since $B_1, B_2, …, B_n$ are mutually exclusive, the occurrence of $A$ means the occurrence of one of $B_1 A, B_2 A, …, B_n A$.
By the addition theorem we get $P(A) = P(B_1 A) + P(B_2 A) + … + P(B_n A)$
And, applying the multiplication theorem to each term of the sum, we get $P(A) = P(B_1) P(A\mid B_1) + P(B_2) P(A\mid B_2) + … + P(B_n) P(A\mid B_n)$.
Example
A store receives products from three factories in proportions of 20%, 30%, and 50%. The first factory produces 10% premium-grade products, the second – 5%, and the third – 20%. What is the probability that a randomly purchased product is premium grade?
- Let event $B_i$ be the purchase of a product from factory $i$.
- Then $P(B_1) = 0.2$, $P(B_2) = 0.3$, and $P(B_3) = 0.5$.
- $A$ is the purchase of a premium-grade product.
- Then $P(A\mid B_1) = 0.1$, $P(A\mid B_2) = 0.05$, and $P(A\mid B_3) = 0.2$.
- By the law of total probability we get $P(A) = \sum_{i = 1}^{3} P(B_i) P(A\mid B_i) = 0.135$
See also
Sources
- Gmurman V.E., Probability Theory and Mathematical Statistics – 9th edition. Moscow: Vysshaya Shkola, 2003.
- Lecture notes on Probability Theory and Mathematical Statistics